1) 24 ⋮ x; 36 ⋮ x ; 150 ⋮ x và x lớn nhất.
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Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1
=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)
`a) 48048 - 48048 : 24 - 24 x 58 `
` = 48048 - 2002- 1392`
` = 46046 - 1392`
` = 44654`
b) 10000 - ( 93120 : 24 - 24 x 57 )
` = 10000 - ( 3880 - 1368)`
` = 10000 - 2512`
` = 7488`
c) 100798 - 9894 : 34 x 23 - 23 =
` = 100798 - 6693 -23`
` = 94082`
`#3107.101107`
\(48\div24-x=1\\ \Rightarrow2-x=1\\ \Rightarrow x=2-1\\ \Rightarrow x=1\)
Vậy, `x = 1`
____
\(124-2\times\left(x+3\right)=24\\ \Rightarrow2\left(x+3\right)=124-24\\ \Rightarrow2\left(x+3\right)=100\\ \Rightarrow x+3=100\div2\\ \Rightarrow x+3=50\\ \Rightarrow x=50-3\\ \Rightarrow x=47\)
Vậy, `x = 47.`
1) 48 : 24 - x = 1 2) 123 - 2 x (x+3) = 24
2-x=1 2 x (x+3) = 123 - 24
x= 2-1 2 x (x+3) = 99
x= 1 x+3 = 99 : 2
x+3 = \(\dfrac{99}{2}\)
x = \(\dfrac{99}{2}\) -3
x = \(\dfrac{93}{2}\)
\(\dfrac{24}{x}:\dfrac{8}{3}=\dfrac{3}{5}\)
\(\dfrac{24}{x}=\dfrac{3}{5}.\dfrac{8}{3}\)
\(\dfrac{24}{x}=\dfrac{8}{5}\)
\(\dfrac{24}{x}=\dfrac{24}{15}\)
=>x=5
Vậy x=5
\(x+3\dfrac{1}{2}+x=24\dfrac{1}{4}\)
\(\left(x+x\right)+3\dfrac{1}{2}=24\dfrac{1}{4}\)
\(x.2+\dfrac{7}{2}=\dfrac{97}{4}\)
\(x.2=\dfrac{97}{4}-\dfrac{7}{2}\)
\(x.2=\dfrac{97}{4}-\dfrac{14}{4}\)
\(x.2=\dfrac{83}{4}\)
\(x=\dfrac{83}{4}:2\)
\(x=\dfrac{83}{4}.\dfrac{1}{2}\)
\(x=\dfrac{83}{8}\)
\(x=10\dfrac{3}{8}\)
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+\left(x^{26}+x^{22}+...+x^2\right)}\)
\(=1-\frac{x^2\left(x^{24}+x^{20}+...+x^4+x^1\right)}{\left(1+x^2\right)\left(x^{24}+2^{20}+...+x^4+1\right)}=1-\frac{x^2}{1+x^2}\)
\(=\frac{1+x^2-x^2}{1+x^2}=\frac{1}{1+x^2}\)
Hoặc cách khác:
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+x^2\left(x^4+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)
a) \(\frac{24}{x}:\frac{8}{3}=\frac{3}{5}\) \(\frac{24}{x}=\frac{3}{5}.\frac{8}{3}\) \(\frac{24}{x}=\frac{8}{5}\) \(x=24.5:8\) \(x=15\)
b) Đề bài sai rồi
Đề đọc khó hiểu quá. Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.
\(\Rightarrow x\inƯCLN\left(24,36,150\right)=6\)
x=6